Does convergence in probability imply weak convergence on an arbitrary metric space? Let $(S, \rho)$ be a metric space and write $\mathcal{S}$ for the corresponding Borel sigma-field. Let $(X_n)_{n = 1}^\infty$ be a sequence of random variables converging in probability to a random variable $X$, by which I mean: for all $\epsilon > 0$ we have as $n \rightarrow \infty$
$$ \mathbb{P}( \rho(X_n, X) > \epsilon ) \rightarrow 0 $$
Is it true that $X_n \Rightarrow X$ in the sense of weak convergence?
 A: It is true.
This is a consequence of the following result, also found in Achim Klenke's book Probability theory (Thm. 13.18 / Cor. 13.19):
Let $ X, X_1, X_2\ldots  $ and $ Y, Y_1,Y_2 \ldots $ random variables with
values in a metric space $ (S,ρ)  $. Assume $ X_n → X $ weakly and $
ρ(X_n,Y_n) → 0   $  in probability. Then $ Y_n → X $ weakly. Considering the special case $X_n = X$ then shows that convergence in probability implies weak convergence.
To show this, note that due to the Portemanteau theorem, we can test weak convergence using bounded Lipschitz continous functions instead of bounded continuous functions.
Next, for any $f$ Lipschitz and bounded, it holds that $\limsup_{n\to\infty}\operatorname{E}\left[\left|f(X_n)-f(Y_n)\right|\right]=0$ (see this question. The assumption of a separable metric space made there is needed merely to ensure that something like $ρ(X,X_n)$ is measurable). Armed with this, we find
\begin{align*}
    \limsup_{n \rightarrow \infty}{|\mathbb{E}[f(Y_n)] - \mathbb{E}[f(X)]  |} 
    \leq \limsup_{n \rightarrow \infty}{|\mathbb{E}[f(X)] - \mathbb{E}[f(X_n)]  | } +
    \limsup_{n \rightarrow \infty}{| \mathbb{E}[f(X_n) - f(Y_n)] |}  = 0,
\end{align*}
hence $ Y_n \rightarrow X $ weakly.
